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Mechanics of Materials: An Integrated Learning System 4th Edition by Timothy A. Philpot Complete Solution Manual

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Mechanics of Materials: An Integrated Learning System 4th Edition Solution Manual by Engineering-Math.org

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Mechanics of Materials: An Integrated Learning Approach 3rd Edition by Timothy Philpot Problem P1.1

A stainless steel tube with an outside diameter of 60 mm and a wall thickness of 5 mm is used as a compression member. If the axial normal stress in the member must be limited to 200 MPa, determine the maximum load P that the member can support.

Solution:

We are given the following values:

\begin{align*}
\text{Outside Diameter}, D &= 60\ \text{mm} \\
\text{Wall Thickness}, t & = 5\ \text{mm} \\
\text{Inside Diameter}, d & = D-2t = 60\ \text{mm}-2\left( 5\ \text{mm} \right) = 50\ \text{mm}\\
\text{Maximum Axial Stress}, \sigma & =200\ \text{MPa} = 200\ \frac{\text{N}}{\text{mm}^2}
\end{align*}

The cross-sectional area of the stainless-steel tube is

\begin{align*}
A & = \frac{\pi}{4}\left( D^2 - d^2 \right) \\
A & = \frac{\pi}{4}\left[ \left( 60\ \text{mm} \right)^2 - \left( 50\ \text{mm} \right)^2 \right] \\
A & = 863.938\ \text{mm}^2
\end{align*}

The normal stress in the tube can be expressed as

\sigma =\frac{P}{A}

The maximum normal stress in the tube must be limited to 200 MPa. Using 200 MPa as the allowable normal stress, rearrange this expression to solve for the maximum load P.

\begin{align*}
P_{max} & = \sigma _{max} A \\
P_{max} & = \left( 200\ \text{MPa} \right)\left( 863.938\ \text{mm}^2 \right)\\
P_{max} & = \left( 200\ \frac{\text{N}}{\text{mm}^2} \right)\left( 863.938\ \text{mm}^2 \right)\\
P_{max} & = 172788\ \text{N} \\
P_{max} & = 172.8\ \text{kN}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}